Standard deviation of frequency distributions

Discrete distribution: σ² = (1/N) Σ fᵢ(xᵢ − x̄)² and σ = √[(1/N) Σ fᵢ(xᵢ − x̄)²], with N = Σfᵢ.

Continuous distribution: replace every class by its mid-point and use the same formula.

Bowling-machine test (made up), x̄ = 6.5: Σfᵢ(xᵢ − 6.5)² = 100, so σ² = 100/40 = 2.5 and σ = √2.5 ≈ 1.58 hits.

Another form avoids subtracting x̄ from every value: σ² = (1/N²)[N Σfᵢxᵢ² − (Σfᵢxᵢ)²], so σ = (1/N)√[N Σfᵢxᵢ² − (Σfᵢxᵢ)²]. It follows from expanding (xᵢ − x̄)² and using Σfᵢxᵢ = N x̄.

Practice match (made up): Σfᵢxᵢ = 1550 and Σfᵢxᵢ² = 58450, so σ² = (50 × 58450 − 1550²)/50² = (2922500 − 2402500)/2500 = 208 and σ = √208 ≈ 14.42 runs.

The same practice-match answer from deviations: Σfᵢ(xᵢ − 31)² = 10400 and 10400/50 = 208.

NCERT's worked results: a discrete distribution with N = 30 has mean 14, variance 45.8 and σ ≈ 6.77; a continuous one with N = 50 has mean 62, variance 201 and σ ≈ 14.18; a third has σ ≈ 6.12.